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2 years ago

What is 12 5/13 as an improper fraction

Mathematics

2 answers:

Maurinko [17]2 years ago

6 0

Answer: 161/13

Step-by-step explanation: 1. Multiply the denominator by The number

2. add the answer from step one to the numerator

3. Write the answer from step 2 over the denominator

aev [14]2 years ago

4 0

161/13 would be the answer

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Nate spent $28 on 2 dvds.at this rate how much would 5 dvds cost him

Anuta_ua [19.1K]

2=28 divide both get 14 . So each cd is 14$. So now 28+28=56 (4 cd's) plus 14 (1 more cd) =5 so 5 is 70.

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3 years ago

Can someone answer this for me? <br><br> Thanks!

MrRa [10]

Hiii :))

Answer is in the attachment.

Identity used :-

1) cosec² α = 1 + cot² α

Trigonometric Ratio :-

1) cot 30° = √3

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Reddam is an expensive school, right?

RainbowSalt2222 ☔

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2 years ago

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The area of a triangle is 30 square centimeters and its base is 10 cm. What is the height in cm of the triangle

Vesnalui [34]

Answer:

The height of the triangle is $6$ cm.

Step-by-step explanation:

----------------------------------------

The formula for solving the area of a triangle is $A=\frac{1}{2}bh$ where $b$ stands for base and $h$ stands for height.

Let's substitute 30 $cm^2$ for the area and 10 for the base.

$30=\frac{1}{2} (10)h$

-------------------->>>>

Now, let's solve to find the height.

$30=\frac{1}{2}(10)h$

Multiply $\frac{1}{2}$ times $10$

$30=5h$

Divide both sides by 5.

$6=h$

-----------------------------------------

Hope this is helpful.

4 0

3 years ago

Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in,and a standard deviation given by

Misha Larkins [42]

Answer:

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with

$\mu$ = 62.5 in

$\sigma$ = 2.2 in

the normal probability density function is given by

$f(x) = \frac{1}{\sqrt{2\pi}2.2}\exp{-\frac{(x-62.5)^{2}}{2(2.2)^{2}}}$ , then

(a) $P(X < 63) = \int\limits_{-\infty}^{63}f(x) dx$ = 0.5899

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking $P(\bar{X} < 63)$ where n = 37. $\bar{X}$ is normally distributed with mean 62.5 in and standard deviation $2.2/\sqrt{37}$ . So, the probability density function is given by

$g(x) = \frac{1}{\sqrt{2\pi}\frac{2.2}{\sqrt{37}}}\exp{-\frac{(x-62.5)^{2}}{2(2.2/\sqrt{37})^{2}}}$ , and

$P(\bar{X} < 63) = \int\limits_{-\infty}^{63}g(x)dx$ = 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

4 0

3 years ago

Find positive integers that satisfy

tankabanditka [31]

9514 1404 393

Answer:

(x, y, z) = (1, 2, 3)

Step-by-step explanation:

The equations that result from reduction to row-echelon form are ...

x = 0.4 +0.2t

y = 5.6 -1.2t

z = t

Then t must have a value 5n+3 for 0 ≤ n < 1. That is, t=3.

x = 0.4 +0.2(3) = 1

y = 5.6 -1.2(3) = 2

z = 3

The integers that satisfy are (x, y, z) = (1, 2, 3).

4 0

3 years ago

Read 2 more answers